# 1.2: Chain Complex of Chain Complexes (Wednesday, January 20) > See phone pic for missed first 10m ## Double Complexes :::{.remark} Consider a double complex: \begin{tikzcd} &&&&&& {C_{p, \cdot}:} \\ &&&& \vdots && \vdots && \vdots \\ \\ && \cdots && {C_{p-1, q+1}} && {C_{p, q+1}} && {C_{p+1, q+1}} && \cdots \\ \\ {C_{\cdot, q}:} && \cdots && {C_{p-1, q}} && {C_{p, q}} && {C_{p+1, q}} && \cdots \\ \\ && \cdots && {C_{p-1, q+1}} && {C_{p, q+1}} && {C_{p+1, q+1}} && \cdots \\ \\ &&&& \vdots && \vdots && \vdots \arrow["{d_{p, q}^h}", from=6-7, to=6-5] \arrow["{d_{p, q}^v}", from=6-7, to=8-7] \arrow["{d_{p, q+1}^v}", from=4-7, to=6-7] \arrow["{d_{p+1, q+1}^v}", from=4-9, to=6-9] \arrow["{d_{p+1, q}^v}", from=6-9, to=8-9] \arrow["{d_{p-1, q}^v}", from=6-5, to=8-5] \arrow["{d_{p-1, q+1}^v}", from=4-5, to=6-5] \arrow[from=8-5, to=10-5] \arrow[from=8-7, to=10-7] \arrow[from=8-9, to=10-9] \arrow["{d_{p+1, q+1}^h}", from=8-9, to=8-7] \arrow["{d_{p+1, q}^h}", from=6-9, to=6-7] \arrow["{d_{p, q+1}^h}", from=8-7, to=8-5] \arrow["{d_{p+1, q+1}^h}"{description}, from=4-9, to=4-7] \arrow["{d_{p, q+1}^h}"{description}, from=4-7, to=4-5] \arrow[from=2-5, to=4-5] \arrow[from=2-7, to=4-7] \arrow[from=2-9, to=4-9] \arrow[from=4-5, to=4-3] \arrow[from=6-5, to=6-3] \arrow[from=8-5, to=8-3] \arrow[from=8-11, to=8-9] \arrow[from=6-11, to=6-9] \arrow[from=4-11, to=4-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) All of the individual rows and columns are chain complexes, where $(d^h)^2 = 0$ and $(d^v)^2 = 0$, and the square anticommute: $d^v d^h + d^h d^v - 0$, so $d^v d^h = -d^h d^v$. This is almost a chain complex of chain complexes, i.e. an element of $\Ch(\Ch \rmod))$. It's useful here to consider lines parallel to the line $y=x$. ::: :::{.definition title="Bounded Complexes"} A double complex $C_{\wait, \wait}$ is **bounded** if and only if there are only finitely many nonzero terms along each constant diagonal $p+q = n$. ::: :::{.example title="?"} A *first quadrant* double complex $\ts{C_{p, q}}_{p, q\geq 0}$ is bounded: note that this can still have infinitely many terms, but each diagonal is finite because each will hit a coordinate axis. ::: :::{.remark title="The sign trick"} The squares anticommute, since the $d^v$ are not chain maps between the horizontal chain complexes. This can be fixed by changing every one out of four signs, defining \[ f_{*, q}: C_{*, q} \to C_{*, q-1} \\ f_{p, q} \da (-1)^p d^v_{p, q}: C_{p,q} \to C_{p, q-1} .\] This yields a new double complex where the signs of each column alternate: % https://q.uiver.app/?q=WzAsNixbMCwwLCJDX3swLCBxfSJdLFsyLDAsIkNfezEsIHF9Il0sWzQsMCwiQ197MiwgcX0iXSxbMCwyLCJDX3swLCBxLTF9Il0sWzIsMiwiQ197MSwgcS0xfSJdLFs0LDIsIkNfezIsIHEtMX0iXSxbMCwzLCJkXnYiXSxbMSw0LCItZF52Il0sWzIsNSwiZF52Il0sWzIsMSwiZF5oIiwxXSxbMSwwLCJkXmgiLDFdLFs1LDQsImReaCIsMV0sWzQsMywiZF5oIiwxXV0= \begin{tikzcd} {C_{0, q}} && {C_{1, q}} && {C_{2, q}} \\ \\ {C_{0, q-1}} && {C_{1, q-1}} && {C_{2, q-1}} \arrow["{d^v}", from=1-1, to=3-1] \arrow["{-d^v}", from=1-3, to=3-3] \arrow["{d^v}", from=1-5, to=3-5] \arrow["{d^h}"{description}, from=1-5, to=1-3] \arrow["{d^h}"{description}, from=1-3, to=1-1] \arrow["{d^h}"{description}, from=3-5, to=3-3] \arrow["{d^h}"{description}, from=3-3, to=3-1] \end{tikzcd} Now the squares commute and $f_{\wait, q}$ are chain maps, so this object is an element of $\Ch(\Ch \rmod)$. ::: ## Total Complexes :::{.remark} Recall that products and coproducts of \(R\dash\)modules coincide when the indexing set is finite. ::: :::{.definition title="Total Complexes"} Given a double complex $C_{\wait, \wait}$, there are two ordinary chain complexes associated to it referred to as **total complexes**: \[ (\Totprod C)_n &\da \prod_{p+q = n} C_{p, q}\\ (\Totsum C)_n &\da \bigoplus_{p+q = n} C_{p, q} .\] Writing $\Tot(C)$ usually refers to the former. The differentials are given by \[ d_{p, q} = d^h + d^v: C_{p, q} \to C_{p-1, q} \oplus C_{p, q-1} ,\] where $C_{p, q} \subseteq \Totsum (C)_n$ and $C_{p-1, q} \oplus C_{p, q-1} \subseteq \Totsum (C)_{n-1}$. Then you extend this to a differential on the entire diagonal by defining $d = \bigoplus_{p, q} d_{p, q}$. ::: :::{.exercise title="?"} Check that $d^2 = 0$, using $d^v d^h + d^h d^v = 0$. ::: :::{.remark} Some notes: - $\Totsum (C) = \Totprod (C)$ when $C$ is bounded. - The total complexes need not exist if $C$ is unbounded: one needs infinite direct products and infinite coproducts to exist in \( \mathcal{C} \). A category admitting these is called **complete** or **cocomplete**.[^dont_exist_ab_cat] [^dont_exist_ab_cat]: Recall that abelian categories are additive and only require *finite* products/coproducts. A counterexample: categories of *finite* abelian groups, where e.g. you can't take infinite sums and stay within the category. ::: ## More Operations :::{.definition title="Truncation below"} Fix $n\in \ZZ$, and define the **$n$th truncation** $\tau_{\geq n}(C)$ by \[ \tau_{\geq n}(C) = \begin{cases} 0 & i < n \\ Z_n & i= n \\ C_i & i > n . \end{cases} .\] Pictorially: \begin{tikzcd} \cdots & 0 & {Z_n} & {C_{n+1}} & {C_{n+2}} & \cdots \arrow[from=1-2, to=1-1] \arrow["{d_n}"', from=1-3, to=1-2] \arrow["{d_{n+1}}"', from=1-4, to=1-3] \arrow["{d_{n+2}}"', from=1-5, to=1-4] \arrow[from=1-6, to=1-5] \end{tikzcd} > [Link to diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJcXGNkb3RzIl0sWzEsMCwiMCJdLFsyLDAsIlpfbiJdLFszLDAsIkNfe24rMX0iXSxbNCwwLCJDX3tuKzJ9Il0sWzUsMCwiXFxjZG90cyJdLFsxLDBdLFsyLDEsImRfbiIsMl0sWzMsMiwiZF97bisxfSIsMl0sWzQsMywiZF97bisyfSIsMl0sWzUsNF1d) This is sometimes call the **good truncation of $C$ below $n$**. ::: :::{.remark} Note that \[ H_i(\tau_{\geq n} C) = \begin{cases} 0 & i < n \\ H_i(C) & i\geq n. \end{cases} .\] ::: :::{.definition title="Truncation above"} We define the quotient complex \[ \tau_{ [Link to Diagram](https://q.uiver.app/?q=WzAsMTYsWzAsMiwiQyJdLFswLDMsIkNbcF0iXSxbMywyLCJDXzAiXSxbMywwLCIwIl0sWzAsMCwiXFx0ZXh0e0RlZ3JlZXN9Il0sWzEsMCwiLXAiXSxbMSwyLCJDX3stcH0iXSxbMywzLCJDX3AiXSxbMSwzLCJDXzAiXSxbMiwyLCJcXGNkb3RzIl0sWzIsMywiXFxjZG90cyJdLFs0LDMsIlxcY2RvdHMiXSxbNCwyLCJcXGNkb3RzIl0sWzUsMiwiQ197cH0iXSxbNSwzLCJDX3sycH0iXSxbNSwwLCJwIl0sWzIsOCwiIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzEzLDcsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) Similarly, if $C$ is a *cochain* complex, we set $C[p]^n \da C^{n-p}$: \begin{tikzcd} {\text{Degrees}} & {-p} && 0 && p \\ \\ C & {C^{-p}} & \cdots & {C^0} & \cdots & {C^p} \\ {C[p]} & {C^0} & \cdots & {C^{-p}} & \cdots & {C^0} \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=3-5, to=3-6] \arrow[from=4-2, to=4-3] \arrow[from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=4-5, to=4-6] \arrow[dashed, from=3-2, to=4-4] \arrow[dashed, from=3-4, to=4-6] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) > Mnemonic: Shift $p$ positions in the same direction as the arrows. In both cases, the differentials are given by the shifted differential $d[p] \da (-1)^p d$. Note that these are not alternating: $p$ is the fixed translation, so this is a constant that changes the signs of all differentials. Thus $H_n(C[p]) = H_{n+p}(C)$ and $H^n(C[p]) = H^{n-p}$. ::: :::{.exercise} Check that if $C^n \da C_{-n}$, then $C[p]^n = C[p]_{-n}$. ::: :::{.remark} We can make translation into a functor $[p]: \Ch \to \Ch$: given $f: C\to D$, define $f[p]: C[p] \to D[p]$ by $f[p]_n \da f_{n+p}$, and a similar definition for cochain complexes changing $p$ to $-p$. :::